zoukankan      html  css  js  c++  java
  • Search in a Binary Search Tree

    Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

    For example, 

    Given the tree:
            4
           / 
          2   7
         / 
        1   3
    
    And the value to search: 2
    

    You should return this subtree:

          2     
         /    
        1   3
    

    In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

    Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

    Approach #1: C++.

    /**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        TreeNode* searchBST(TreeNode* root, int val) {
            while (root != nullptr && root->val != val) {
                root = (root->val > val) ? root->left : root->right;
            }
            return root;
        }
    };
    

      

    Approach #2: Java.

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode searchBST(TreeNode root, int val) {
            if (root == null || root.val == val) return root;
            return root.val > val ? searchBST(root.left, val) : searchBST(root.right, val);
        }
    }
    

      

    Approach #3: Python.

    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def searchBST(self, root, val):
            """
            :type root: TreeNode
            :type val: int
            :rtype: TreeNode
            """
            if root and root.val > val:
                return self.searchBST(root.left, val)
            if root and root.val < val:
                return self.searchBST(root.right, val)
            return root
            
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    Spring Boot 的单元测试和集成测试
    Containers vs Serverless:你选择谁,何时选择?
    Java13新特性
    Java中创建对象的5种方法
    最好的重试是指数后退和抖动
    杂谈:面向微服务的体系结构评审中需要问的三个问题
    使用Quarkus在Openshift上构建微服务的快速指南
    Java EE—最轻量级的企业框架?
    AQS机制
    JVM-内存模型
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10011498.html
Copyright © 2011-2022 走看看